Math Is Fun Forum

Got it  - makes sense -

1. I made a mistake assuming density function is 0 beyond x = 250- it is clearly 250

2. Clearly, if the median for the exponential density function works out to be less than the benefit limit, the limit doesnt change the median

Thanks bobbym! I got 2 more problems for you - Stay tuned!

not sure if it sounds like binomial to me (not disagreeing - just not sure) - for the following reasons -

experiment following binomial distribution should have -

1. n identical trials
2. each should have only 2 outcomes - success and failure
3. probability of success on each trial should stay the same
4. all trials should be independent of each other

IT COULD WELL BE JUST ME - as with these problems, it just seems one needs to have the 'eye' to fit a problem to an established distribution!

Yep - they make sense - thanks for all the guidance and encouragement! I will keep trying.

I will let you know if I get stuck again.

- getback0

First up! THANKS A LOT - for the prompt response! I need to be more regular at this stuff.

I have a few questions though -

1.  For Q #4:
=========

In the solution when we substitute

; we are assuming A and B to be independent events - data that the problem has not provided, correct?

2.  For Q #3:
=========

In the solution again, we have used -

which suggests we use -

for the probability of a certain number of claims

Do we?? If so, this is the probability of a number of injury claims, is it not??

3.  For Q #2:
=========

When they say the proportion of people choosing coverage A is

  , how does that suggest

? You clearly got it right - why am I not getting it even when someone's laying the solution out for me??????????

4.  For Q #1:
==========

Ok ... let me just say .... YOU ARE AN AUTHORITY on this subject!!!

Could you help me understand what's wrong with my thought process? I totally see your solution. I like to verbalize the experiment and then translate that into a "counting" problem.

From your solution, it seems you translated this to -

X: Pick 2 different ranks, and pick any 2 cards from a total of 4 for each rank ->

Y: Pick another rank, and pick any 1 card from a total of 4 for each rank        ->

n(E) = X * Y

What is wrong with my line of reasoning?

Alright! Alright! Alright!

I think I got it!!! until someone proves this wrong ...

The number of unique partitions (k) of n objects into p groups of m objects each (where n = m*p) is -

Hence, for the example, number of uniqe partitions of a group of 4 people into 2 groups of 2 each is -

C(4,2) * C(2,2) / P(2,2) = 6*1/2 = 3

This is verified in the example on the original post.

Here is why I think "k" should have the value from (1.1) -

1. The relation in (1.1) without the P(p,p) or p! is the familiar result for the outcome of an experiment that deals with picking m objects of n successively for p times when n = m*p

2. However, in this result, the same combination of objects appear multiple times - once for each of the p groups. In our example, [ (1,2), (3,4) ] is a partition.

This is different from [ (3,4), (1,2) ] if you define Groups - A and B - so, the C(4,2) * C(2,2) is trying to count both partitions. But if you consider them to be the same partition, just arranged in a different order, then this is just a rearrangement of the 2 partitions - in the general case, the p partitions would be rearranged p! times - inflating the results by a factor p!

3. Consequently, the final result needs to be (1/p!) times the outcome from C(n, m) * C(n - p, m) * C(n - 2p, m) * ... * C(m, m)

I tried verifying this with dividing 6 objects into groups of 2 objects -

conventionally this should have been C(6, 2) * C(4, 2)  * C(2, 2) = 15 * 6 * 1 = 90

However, if you were to actually list the combinations down, you will see we start repeating combinations around in a different orde - rightly so, if you are looking for number of ways to group them in Groups A, B and C. However, if you are simply looking to split them in groups of 2, you will need to weed out 3! - 1 redundant counts - for each combination. If you do that, you should end with 15 unique partitions - which is the answer from 1.1

Thoughts bobby?

Leon

may be not - I dont know any more

Hello Bobbym,

I meant to get back sooner - first up, thanks for the detailed explanation! It makes sense to me - I was just struggling conceptually in that for E2, I felt I could -

1.  Determine number of ways to split 4 people in groups of 2 IN ANY ORDER
2.  Determine number of ways to pick 2 numbers from 1 - 10 and assign them in a specific order to all the combinations in #1

Since #1 above had no sense of an order, I used C(4,2)*C(2,2).

Also, since #2 above sounded like it had order involved, I used P(10,2).

... giving me C(4,2)*C(2,2)*P(10,2) = 540.

However, i now realize that using C(4,2) [number of combinations of 4 objects taken 2 at a time] would cause repititions in the splits I make. For example,

P-1 P-2         P-3 P-4
P-1 P-3         P-2 P-4
P-1 P-4         P-2 P-3
================ (=3) this right here is all possible partitions of 4 people in groups of 2. However, C(4,2) continues ...
P-2 P-3         P-1 P-4
P-2 P-4         P-1 P-3
P-3 P-4         P-1 P-2
================ (=6) all these partitions in the second group here are dups of the one above

I realize now that #1 above was really the unique number of partitions I could make (each partition is a group of combinations). So, in this case (1/2)*C(4,2) would the number of outcomes of #1 and hence -

(1/2)*C(4,2)*P(10,2) = 270

Now that sounds like a desperate attempt to match the right answer - however, I would feel better if you could weigh in on the following hypothesis -

The total number of unique partitions (k) of n objects when it is divided in p groups of m objects each for a given partition is -

k = (1/p)*C(n,m)

As always, thanks a lot for the guidance!

Leon